How do you find the value of the other five trigonometric functions, given tan x=4/11, sec x< 0?

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Answer 1 · 2015-10-10T12:58:25

#sinx=-2/sqrt15, cosx=-sqrt(11/15)#

#tan x=4/11, cotx=11/4#

#secx=-sqrt(15/11), cscx=-sqrt15/2#

#sec^2x=tan^2x+1#

#sec^2x=4/11+1=4/11+11/11=15/11#

#secx=sqrt(15/11) vv secx=-sqrt(15/11)#

So, #secx=-sqrt(15/11)#

#cosx=1/secx#

#cosx=-sqrt(11/15)#

#sin^2x+cos^2x=1 => sin^2x=1-cos^2x#

#sin^2x=1-11/15=4/15#

#sinx=-2/sqrt15 vv sinx=2/sqrt15#

Since #tanx=4/11=sinx/cosx# and #cosx<0# it must hold #sinx<0#

So, #sinx=-2/sqrt15#

#cotx=1/tanx=11/4#

#cscx=1/sinx=-sqrt15/2#

Finally:

#sinx=-2/sqrt15, cosx=-sqrt(11/15)#

#tan x=4/11, cotx=11/4#

#secx=-sqrt(15/11), cscx=-sqrt15/2#